The type of the pointer [duplicate]

Question

I am new to C++ and as I was reading MIT lecture note about pointer I recognize something strange:

Pointers are just variables storing integers – but those integers happen to be memory ad­dresses, usually addresses of other variables. A pointer that stores the address of some variable x is said to point to x. We can access the value of x by dereferencing the pointer.

well and also I find that the pointer can have a type:

int *pointer ; 
char * pointer ; //example 

well it just said it's an int that hold an address why give it the same type as the thing it point at if it's just hold a reference to it not an actual value ?

Answer 1

Maybe you have an array of int. and "accidently" ... an array variable is a pointer to the first component of that array. With the knowledge of having a pointer being of type int, the compiler knows how far it must jump to the next data.

int a[4] = { 0, 1, 2, 3 };

// Dereferenced pointers to an int with additonal offset of size int n times
printf("%i\n", *a);
printf("%i\n", *(a+1));
printf("%i\n", *(a+2));
printf("%i\n", *(a+3));

// Equivalent to using the array as usual
printf("%i\n", a[0]);
printf("%i\n", a[1]);
printf("%i\n", a[2]);
printf("%i\n", a[3]);

This is an advanced example, but it is one of the seldom uses for an int pointer. Most often you will have pointers to objects, structs or arrays - seldom to the value types. But in some implementations of algorithms, pointers to value types can be usefull for sorting, searching etc. in arrays